As an example of the manipulation of an electron beam, consider an electron trav
ID: 1338524 • Letter: A
Question
As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the x axis in the xy plane with initial velocity vi = vii. As it passes through the region x = 0 to x = d, the electron experiences acceleration a = axi + ayj, where ax and ay are constants. For the case vi = 1.88 10^7 m/s, ax = 8.64 10^14 m/s2, and ay = 1.80 10^15 m/s2, determine the following, at x = d = 0.0100 m.
(a) the position of the electron yf = m
(b) the velocity of the electron vf = m/s i + m/s j
(c) the speed of the electron |vf| = m/s (
d) the direction of travel of the electron (i.e. the angle between its velocity and the x axis) = °
Explanation / Answer
Initially we have Velocity in X direction only that is Vi = 1.88*107 m/s
So after covering distance d in x direction the velocity of particle in x direction
V2 = u2 +2axd
VX = 1.925*107 m/s
Now we will calcualte time
V = u+at
t = V-u /a = (1.925 - 1.88)*107 / (8.64*1014) = 5.208*10-10 s
Now we have to calcualte the distance covered in the Y direction
Since we have zero initial velocity in Y direction
VY = u+aYt = 0+(1.8*1015*5.208*10-10) = 9.3744*105 m/s
Distance covered in the Y direction
Y = (1/2)at2 = (1/2)(1.8*1015)(5.208*10-10)2 = 0.000244 m
(a) Yf = 0.01 i + 0.000244 j
(b)Vf = 1.925*107 i + 9.3744*105 j
(c) Speed of the electron V = (VX2 + Vy2)1/2
V = 1.927*107 m/s
(d) tan(angle) = VY/VX = 9.3744*105 /(1.925*107)
angle = tan-1(VY/VX)
angle = 2.78 degree
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